Question

The α-decay of an excited 2⁻ state in ¹⁶O to the ground 0⁺ state of ¹²C is found to have a width Tα ≅ 1.0 × 10⁻¹⁰ eV. Explain why this decay indicates a parity-violating potential.

Answer 1

The initial state ¹⁶O^∗ has odd parity, while the parity of ¹²C is even and so also that of α. If α is emitted with l = 0 the parity of the final state is even. Clearly the decay goes through a weak interaction because parity is violated and the observed width is consistent with a weak decay. On the other hand the width of the electro-magnetic decay ¹⁶O^∗ →¹⁶O + γ, is 3 × 10⁻³eV.